GST95
20+ Year Contributor
- 222
- 2
- Jun 23, 2002
-
Bristol,
Indiana
The page I linked to above has all the answers you're looking for, including a complete explanation on how the 50% number is derived and how to calculate the gasoline equivalent flow rate of injectors on E85.
For 1050cc injectors, you can expect the gasoline equivalent of about 700cc:
1050 * 0.67 = 704
But....people generally find that they can run leaner on E85 than they could on gasoline. So that sorta kinda raises that flow rate calculation. Sorta.
For example, if you find that you can run 11.5:1 on E85 where you could only run 10.5:1 on gasoline (I'm normalizing lambda here to be gasoline-based just for simplicity), then that 704 is sorta like 770:
703 * 11.5 / 10.5 = 770
But that's really a goofy way of looking at it, IMO. It's valid, just a little strange and imprecise. It's better to just multiply by 0.67 and then if you find you can lean the mixture out more on E85, great...that's even more head room to play with at WOT. But the normal stoichiometric closed loop operation is still going to be based on that 0.67 factor.
Thomas Dorris
I dont think its a goofy way of looking at it. When people explain it as "you need 30% more to run e85" they take the ability to run leaner into the mix as well. Using your numbers and 12:1(gas scalle) vs 10.5:1 gives you that 30% number. 803 gas = 1050 e85 at WOT.
I also dont really look at it as being able to run leaner with e85, but more that you dont have to run so rich with e85 like you do with gas to keep gas from knocking. 10.5:1 is outside of the rich-lean power window of gas, but 12:1(gas scale) for e85 is.